Question

Need 2-8

- CHAPTER QUESTIONS Reliability 245 1. Definere blir ne reliability in your own words. Describe the key elements and why they are important 2. Describe the three phases of the life syde curve. Drew the curve and label it in detail (the axes, phases, type of product failure, etc.). 3. Determine the failure rate for the following You have tested circuit boards for failures during a 300-hour continuous use test. Four of the 25 boards tailed. The first board failed in No hours, the second tailed in 150 hours, the third failed in 350 hours, the fourth in 465 hour. The otherboards completed the 500-hour test satisfactorily. What is the mean life of the product! 4. Determine the fire rate foto-hour test of 12 inms where 2 items fail at 45 and 72 hours, respectively. What is the mean life of the product? 5. A power station has installed 10 new generators to provide electricity for a local metropolitan area. In the past year (8,760 hours), two of those generators have failed, one at 2.460 hours and one at 5,962 hours. It took five days, working 24 hours a day, to repair cach generator. Using one year as the test period, what is the mean time between failures for these generators? Given the repair information, what is the availability of all 10 generators? 6. Why is a parallel system more reliable than a system in series? 7. Given the system below, what is the system reliability? RC -0.94 PA=0.99 RB = 0.98 RE -0.98 D=0.95 8. Given the diagram of the system below, what is the system reliability? 0.89 0.95 0.99 0.95

Answer 1

V . Souhr continow tests mean lefe = sot 150+ 350+ 465 + 4x500 18 - 380.625hr de facuse satea no. of foulule time da 4 = 8x10 s hr. 500 hr for 90-hrtest, failure 4 failure sate - 2 = 0.022 (ar 90 mean lefe = = 84.75 hr 45+72+ lox 90 12 (9) \о шо ам. давед. Testing Period = 1 year = 876ohr & generator fooled Mean time blu failure= 8760 = 4380kr 8 gerelatos coill be available for whole year prese generator coill be awaualle fol pero 8760 - (Quxs) 00 = eucetto 8640 her

Coelu allel mose that system beliable selles, as Reliability in series cap So any of them Reliability in Parallel Rs = RiXR2XRS fael, whole system fail Rs= Rit Rax (Prob.of failure of 1) So we can see that in parallel facture of I componeert, does not fail the whole systems, That's why, Paralle System more beliable than Series Rc RA 1o.gg f fo.98 (RBURE] RE/0.98 10.95 RD on 10.9996 [0.997] = 0.98 +0.02X0.48 = 0.9996 (RCORD 20.94 +0.06x0.95 Rrod. R system - 0.99x0.9996x 0.997 = 0.9866

Kluring 1889 Fo.897 1898 - 18.95 5.99 10.99 H 10.89 (0.956 (143] 0.89 +0.1180.89 = 0.9879 -0.99748.98798.98] 10.99757 1526 = 0.99x0.9879% 0.98x 0.9975 0.95+0.05XO9T Rsystem 0.9560 = 0.9975 le curve (Bath Tub cerve) - Craphical sep. of life time of population of a product - at show the relative facture late of an enthe population 1 oft Intant mortality wear out I = Infant mortality here failure late dec with time. failure kate Usefellefestas operated lefe/age Ind Phase = failure sate semain const useful libe I as most signifuant Period fol reliabite forediction and phase wear out stage failure eate increase as product begins to weal